Maximize weekly $ given changing price for parts & wholes - algorithm

Question

I have a dollar amount I wish to maximize my spend on products to purchase for a given list of projects.
For example $60. I cannot spend more than $60 a week but can hold over any left over amount. So, if I spend $58 I get to spend $62 the next week. If I purchase a product on week 1 I can use the left over amount on week 2 thereby not needing to re-purchase the same item.
The solution needs to be generic so that I can maximize between a lot of products and a lot of projects given that fixed dollar amount per week.
I need to generate a report with the list of products to purchase and the list of projects to do for that week. The prices get updates weekly so I will need to recalculate the max spend weekly (meaning forecasting is not really part of the solution) and I need to reuse the amount from the products/inventory already purchased.
I have all the data and there aren't any unknown variables. I just need to be able to figure out how to maximize what to purchase given parts and wholes under a fixed dollar amount with history.
To make it more concrete through example (although abstracted from my real data):
I have a database of products (12.5 K different ones) and their corresponding prices. I also have a list of fixed list projects (let's say 2500) I wish to do with those products. For each project I have the corresponding products needed for each. Each project takes a different amount of product. Projects can have overlapping or unique products per project.
So for example:
project 1 is building my model airplanes;
project 2 is fixing my picture frames;
project 3 is building a bird house;
etc.
Project 1 may need:
glue (1oz)
paint (3 qt)
balsam wood (2 lbs)
Project 2 may need:
glue (2 oz)
nails (10 count)
project 3 may need:
glue (10 oz)
paint (5 qts)
nails (40 count)
wood balsam (3 lbs)
wood pine (50 lbs)
Products:
Glue 4oz - $10
Paint 3qts - $30
Nails 15 count - $7
Wood Balsam 8 pounds - $12
Wood Pine 12 pounds - $8
For example, if I buy a bottle of glue (4 oz) at $10 I can use it for my airplanes and my picture frames but not my bird house. I need to do an exhaustive analysis of all products and all projects weekly given my dollar amount to spend since the prices change (sales/demand/etc.).
How can I best spend the $60 to do as many projects as possible in a given week? Week 2 I get a new $60 to spend, (most likely have leftover) money, and have product (such as glue) left over from the week before?
Any python code / projects that do something similar or exactly this already that I may be able to import and modify for my needs?
Any help in terms of algorithms, sample code, full solution!?!, ideas, etc. would be appreciated...
Thanks in advance!!! (FYI: his is for a personal project.)

Answer 1

This is a problem which is very well suited to be tackled with mathematical programming. With mathematical optimization you can optimize variables (e. g. a variable says if a project is conducted at some point) with an objective like the numbers of projects conducted while also considering a set of constraints. For Python there are several free libraries for optimization of mathematical programs, I will show how to get started with your problem using PuLP. Please note that free software for these kind of problems usually performs way worse than commercial one, which can be very expensive. For small or easy problems the free software suffices though.
To get started:
easy_install pulp
Now, import pulp and as a little help itertools.product. There are many ways to represent your data, I choose to declare some ranges which serve as index sets. So r = 0 would be glue and p = 0 build a model air plane. The number of time periods you have to choose. With 4 time periods all projects can be conducted eventually.
from pulp import *
from itertools import product
R = range(5) # Resources
P = range(3) # Projects
T = range(4) # Periods
Your parameters could be represented as follows. project_consumption[0, 0] expresses that project 0 needs 1/4 of material 0 (glue) to be conducted.
resource_prices = (10, 30, 7, 12, 8) # Price per unit of resource
# Needed percentage of resource r for project p
project_consumption = {(0, 0): 1/4, (0, 1): 3/3, (0, 2): 0/15, (0, 3): 2/8, (0, 4): 0/12,
(1, 0): 2/4, (1, 1): 0/3, (1, 2): 10/15, (1, 3): 0/8, (1, 4): 0/12,
(2, 0): 10/4, (2, 1): 5/3, (2, 2): 40/15, (2, 3): 3/8, (2, 4): 50/12,}
budget = 60
Next, we declare our problem formulation. We want to maximize the number of projects, so we declare the sense LpMaximize. The decision variables are declared next:
planned_project[p, t]: 1 if project p is conducted in period t, else 0
stocked_material[r, t]: Amount of material r which is on stock in t
consumption_material[r, t]: Amount of r that is consumed in period t
purchase_material[r, t]: Amount of r purchased in t
budget[t]: Money balance in t
Declare our problem:
m = LpProblem("Project planning", LpMaximize)
planned_project = LpVariable.dicts('pp', product(P, T), lowBound = 0, upBound = 1, cat = LpInteger)
stocked_material = LpVariable.dicts('ms', product(R, T), lowBound = 0)
consumption_material = LpVariable.dicts('cm', product(R, T), lowBound = 0)
purchase_material = LpVariable.dicts('pm', product(R, T), lowBound = 0, cat = LpInteger)
budget = LpVariable.dicts('b', T, lowBound = 0)
Our objective is added to the problem as follows. I multiply every variable with (len(T) - t), that means a project is worth more early rather than later.
m += lpSum((len(T) - t) * planned_project[p, t] for p in P for t in T)
Now we can restrict the values of our decision variables by adding the necessary constraints. The first constraint restricts our material stock to the difference of purchased and consumed materials.
for r in R:
for t in T:
if t != 0:
m += stocked_material[r, t] == stocked_material[r, t-1] + purchase_material[r, t] - consumption_material[r, t]
else:
m += stocked_material[r, t] == 0 + purchase_material[r, 0] - consumption_material[r, 0]
The second constraint makes sure that the correct amount of materials is consumed for the projects conducted in each period.
for r in R:
for t in T:
m += lpSum([project_consumption[p, r] * planned_project[p, t] for p in P]) <= consumption_material[r, t]
The third constraint ensures that we do not spend more than our budget, additionally the leftover amount can be used in future periods.
for t in T:
if t > 0:
m += budget[t] == budget[t-1] + 60 - lpSum([resource_prices[r] * purchase_material[r, t] for r in R])
else:
m += budget[0] == 60 - lpSum([resource_prices[r] * purchase_material[r, 0] for r in R])
Finally, each project shall only be carried out once.
for p in P:
m += lpSum([planned_project[p, t] for t in T]) <= 1
We can optimize our problem by calling:
m.solve()
After optimization we can access each optimal decision variable value with its .value() method. To print some useful information about our optimal plan of action:
for (p, t), var in planned_project.items():
if var.value() == 1:
print("Project {} is conducted in period {}".format(p, t))
for t, var in budget.items():
print("At time {} we have a balance of {} $".format(t, var.value()))
for (r, t), var in purchase_material.items():
if var.value() > 0:
print("At time {}, we purchase {} of material {}.".format(t, var.value(), r))
Output:
Project 0 is conducted in period 0
Project 2 is conducted in period 3
Project 1 is conducted in period 0
At time 0 we have a balance of 1.0 $
At time 1 we have a balance of 1.0 $
At time 2 we have a balance of 61.0 $
At time 3 we have a balance of 0.0 $
At time 0, we purchase 1.0 of material 0.
At time 3, we purchase 1.0 of material 3.
At time 0, we purchase 1.0 of material 3.
At time 1, we purchase 2.0 of material 1.
At time 0, we purchase 1.0 of material 2.
At time 3, we purchase 3.0 of material 2.
At time 3, we purchase 6.0 of material 4.
At time 0, we purchase 1.0 of material 1.
At time 3, we purchase 4.0 of material 0.
Note in the solution we purchase 6 units of material 4 (6*12 wood pine) at time 3. We never really use that much but the solution is still considered optimal, since we do not have budget in our objective and it does not impact the amounts of project we can do if we buy more or less. So there are multiple optimal solutions. As a multi-criteria optimization problem, you could use Big-M values to also minimize budget utilization in the objective.
I hope this gets you started for your problem. You can find countless resources and examples for mathematical programming on the internet.

Answer 2

Something like this could work, creating a list of materials for each project, a list of projects, and dictionary of prices. Each call of compareAll() would show the cheapest project in the list. You could also add a loop which removes the cheapest project from the list and adds it to a to-do list each time it runs, so that the next run finds the next cheapest.
p1 = ["glue","wood","nails"]
p2 = ["screws","wood"]
p3 = ["screws","wood","glue","nails"]
projects = [p1,p2,p3]
prices = {"glue":1,"wood":4,"nails":2,"screws":1}
def check(project,prices):
for i in project:
iPrice = 0
projectTotal = 0
total = 0
for n in prices:
if(i == n):
iPrice = prices[n]
total = total + iPrice
print("Total: " + str(total))
return total
def compareAll(projectList):
best = 100 #Or some other number which exceeds your budget
for i in projectList:
if (check(i,prices) < best):
best = check(i,prices)